Given the function below and point P, what is the direction of the steepest ascent?

Question

Given the function below and point P, what is the direction of the steepest ascent?

 

f(x)=x^2−4y^2−9

Answer 1

It is just the unit vector in the direction of the gradient.

 

∇f(P) = ⟨2x , -8y⟩

 

||∇f(P)|| = √[(2x)² + (-8y)²] = 2√(x² + 16y²)

 

Thus we have

 

u = ∇f(P) / ||∇f(P)|| = ⟨x/√(x² + 16y²) , -4y/√(x² + 16y²)⟩.